Can anyone help me to complete my course

[mirzajee]

Hi i am taking online course i passed almost all just 2 books left i did few questions but sorry some questions are tricky as english is my 2nd language so i am confused . i will be thank full to any who helps me my email is Removed, please use the private message feature to exchange emails, Bob

 

[George Stolz]

Mirzajee, what's your first language? What are you studying, exactly? How long are you needing help?

Giving more information improves your chances of getting a reply.

 

[mirzajee]

thankyou George

thankyou George

Mr George i am taking onlione exam from pennfoster for electrician diploma my 1st language is urdu i have completed 44 books only 2 left my grade is 81% but now i am so confused about this NEC coz NEC is so much complicated not easy to find right answer

 

[charlie b]

I agree that the NEC is confusing. We would probably all agree. But we cannot answer a question that you have not asked. What kind of help are you looking for? A tutor? If you do not understand how to do a problem that is in your book, then post that question, tell us what you have tried, and tell us what part has you confused.

 

[mirzajee]

Oven Calculation

Oven Calculation

Mr Charlie and others
question is 15 Kw operates on 240 volts . What is demand load
Answer as i did
15kw - 12kw =8kw 3X5% =1.2 kw+8kw =9.2kw =9200watts
9200watts/240 is 38.33 amps
neutral 70% of 9200watts =6440 watts 6440/240=26.83 amps 26.83+38.33=65.16 amps is demand load is this correct?
Q 2 what size of TW wire for this circuit ?
6awg?

 

[roger]

Mirzajee, the demand would stop at your 9200 watts, the neutral would not add to this as far as demand.

The ungrounded branch circuit conductors would indeed be 9200 / 240 = 38.33 amps as you have correctly calculated, This would take # 8 conductors.

The neutral would be 70% of 9200 which would be 6440 / 240 = 26.83 amps as you also correctly calculated and would need a # 10 conductor.

So you would have a demand of 9200 watts.

You would need a 40 amp branch circuit consisting of (2) #8 TW conductors and (1) #10 TW neutral conductor. (Note: the equipment grounding conductor is not part of the question)

Roger

 

[mirzajee]

thank you rogers

thank you rogers

thankyou mr rogers i am happy that my calculation was not wrong just i missunderstood question

 

[George Stolz]

Mirzajee, your calculation was wrong:

neutral 70% of 9200watts =6440 watts 6440/240=26.83
26.83+38.33=65.16 amps is demand load

You were correct, and then added a few extra steps to get to a wrong answer. Do you understand?

 

[mirzajee]

Question

Question

1 8Kw cooking unit
2 6Kw Wallmounted owens
8kw
2x 6Kw 12Kw=20kw -12kw= 8kwx5=40%=3.2kw total 11.2kw= 11200watts
what is tw size ?is 6AWG?

 

[mirzajee]

Commercial Load Calculation

Commercial Load Calculation

Question:- In commercial kitchen three 3 kw ovens 1 20kw water heater and one 3 kw deepfryer? how to calculate and what section in nec 2005 i can find regulation? what is demand load?

 

[JohnJ0906]

220.56 kitchen

 

[George Stolz]

So know, knowing the section, what answer do you come up with, Mirzajee?

 

[mirzajee]

NEC is a messup

NEC is a messup

Mr George NEC is big Messup for me or maybe i am new so it seems me to confusing

 

[George Stolz]

Don't get discouraged.

Question:- In commercial kitchen three 3 kw ovens 1 20kw water heater and one 3 kw deepfryer? how to calculate and what section in nec 2005 i can find regulation? what is demand load?

• three 3 kw ovens
• 1 20kw water heater
• one 3 kw deepfryer?

We have three ovens and a deep-fryer that are definutely kitchen equipment. I would look closer at the section to be sure water heaters are included in figuring out the demand load for an "other than dwelling unit" kitchen.

220.56 Kitchen Equipment — Other Than Dwelling Unit(s). It shall be permissible to calculate the load for commercial electric cooking equipment, dishwasher booster heaters, water heaters, and other kitchen equipment in accordance with Table 220.56. These demand factors shall be applied to all equipment that has either thermostatic control or intermittent use as kitchen equipment. These demand factors shall not apply to space-heating, ventilating, or air-conditioning equipment.
However, in no case shall the feeder or service calculated load be less than the sum of the largest two kitchen equipment loads.

So, water heaters are included. So, we have five appliances total to demand.

By Table 220.56, we are allowed to multiply the nameplate ratings by 70% to find our demand load. So, what do you get?